# How to make $\lim_{x \to 0} (1-2x)^{1/x}$ into $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form for L'Hospital's Rule? [duplicate]

I am trying to figure out how to make $$\lim_{x \to 0} (1-2x)^{1/x}$$ into $$\frac{\infty}{\infty}$$ or $$\frac{0}{0}$$ form for L'Hospital's Rule.

Using Desmos.com I have found that $$(1-2x)^{1/x} \neq 1-2^{1/x}x^{1/x}$$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.

## marked as duplicate by Xander Henderson, LuminousNutria, Community♦Apr 8 at 20:55

• Take $\log$ both side – Mann Apr 8 at 19:28
• math.stackexchange.com/questions/2493771/… – Xander Henderson Apr 8 at 19:31
• It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate. – Xander Henderson Apr 8 at 20:48

A suggestion:

Calculate the limit of the logarithm first: $$\lim_{x\to0}\ln\left(1-2x\right)^{1/x}=\lim_{x\to 0}\frac{\ln(1-2x)}x.$$

If you find the limit $$\ell$$ for the log, the limit of the expression is $$\mathrm e^{\ell}$$.

• If I differentiate both sides I can get $\frac{-2}{1-2x}$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $\frac{1}{e^2}$. Is there something else I have to do? – LuminousNutria Apr 8 at 19:43
• $-2$ is the limit of the log. Hence the limit of the expression is … – Bernard Apr 8 at 19:47
• I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up. – LuminousNutria Apr 8 at 19:49
• $(1-2x)^{1/x}=\mathrm e^{\frac 1x\ln(1-2x)}$ since the exponential fiunction is the inverse function of the natural logarithm.. – Bernard Apr 8 at 19:54
• Oh! I see! I get it now $\lim_{x \to 0} (1-2x)^{1/x} = e^{\lim_{x \to 0} \frac{\ln(1-2x)}{x}}$ The exponent of $e$ becomes $-2$. – LuminousNutria Apr 8 at 20:00